Moment of inertia of human body with limbs at an angle

[sunsetpearl]

TL;DR Summary

I'm trying to find the moment of inertia of a human body with (say) the upper arm at angle alpha to the torso and the lower arm an another angle beta to the upper arm

I'm trying to find the moment of inertia of a human body with (say) the upper arm at angle alpha to the torso and the lower arm at another angle beta, where beta is the angle between the lower and upper arms.

I model the torso as a cylinder of radius R, mass M and the parts of the arm as cylinders with radii r1 and r2, mass m1,m2. The upper arm (cylinder 1) is attached to the outside of the large cylinder.
I can easily find the inertia of a spinning human if the arm is outstretched or down by their side but I'm having trouble working it out when the different parts of the arm are at different angles.

I first modelled the arms as cylinders of length L1, L2 - if the angle between axis and cylinder 1 is a, I believe the inertia of the upper arm is m1L1^2sin^2(a)/3 but I'm not sure how to do the same with the lower arm. I tried to use the parallel axis theorem, taking the component of the limb parallel to the axis of rotation. Doing this gives me (L2(cos(pi-a-b)m2r2^2)/2 + m2(L1sin(a))^2 but I'm not sure if I'm going completely wrong here.

For context, I'm trying to figure out how holding your arms in different positions will affect an ice skater while spinning.

 

[erobz]

TL;DR Summary: I'm trying to find the moment of inertia of a human body with (say) the upper arm at angle alpha to the torso and the lower arm an another angle beta to the upper arm

I'm trying to find the moment of inertia of a human body with (say) the upper arm at angle alpha to the torso and the lower arm at another angle beta, where beta is the angle between the lower and upper arms.

I model the torso as a cylinder of radius R, mass M and the parts of the arm as cylinders with radii r1 and r2, mass m1,m2. The upper arm (cylinder 1) is attached to the outside of the large cylinder.
I can easily find the inertia of a spinning human if the arm is outstretched or down by their side but I'm having trouble working it out when the different parts of the arm are at different angles.

I first modelled the arms as cylinders of length L1, L2 - if the angle between axis and cylinder 1 is a, I believe the inertia of the upper arm is m1L1^2sin^2(a)/3 but I'm not sure how to do the same with the lower arm. I tried to use the parallel axis theorem, taking the component of the limb parallel to the axis of rotation. Doing this gives me (L2(cos(pi-a-b)m2r2^2)/2 + m2(L1sin(a))^2 but I'm not sure if I'm going completely wrong here.

For context, I'm trying to figure out how holding your arms in different positions will affect an ice skater while spinning.

Can you draw a picture of the setup with the angles and relevant dimensions indicated.

 

[Baluncore]

Welcome to PF.

For context, I'm trying to figure out how holding your arms in different positions will affect an ice skater while spinning.

Can we assume that you are the ice skater, that they are your arms, and that you are spinning?

 

[sunsetpearl]

Welcome to PF.

Can we assume that you are the ice skater, that they are your arms, and that you are spinning?

Yes! Sorry for not clarifying fully.

 

[sunsetpearl]

Can you draw a picture of the setup with the angles and relevant dimensions indicated.

I didn't include both arms in the diagram, but I assume the other arm would also be at some angle to the torso, not necessarily the same angle as the arm pictured.

 

[A.T.]

I can easily find the inertia of a spinning human if the arm is outstretched or down by their side but I'm having trouble working it out when the different parts of the arm are at different angles.

Use linear algebra instead of trigonometry:
https://scienceworld.wolfram.com/physics/MomentofInertiaCylinder.html
https://en.wikipedia.org/wiki/Moment_of_inertia#Inertia_tensor

 

[erobz]

If you do wish to do the trig, I think you have the correct concept. (Parallel axis theorem, with thin rods) but you need to check extremes of the equation you develop to see over what angles it holds and what angles it may break down.

Write the equation for $m_2$ using the diagram. The red arrow is the distance $x$ of the parallel axis from the axis of rotation of rod $m_2$.

 

[sunsetpearl]

Use linear algebra instead of trigonometry:
https://scienceworld.wolfram.com/physics/MomentofInertiaCylinder.html
https://en.wikipedia.org/wiki/Moment_of_inertia#Inertia_tensor

Could you expand a bit please? I would like to represent it in this way but I thought it might be simpler to do the trig first and understand that.

I've thought about how I'd go about it but I'm quite confused. I guess you might choose the axes as the axis through the center of the cylinder and through the centers of the two rods? So the diagonal components would be easy to calculate but then it seems like the off diagonals would be rather complicated. And adding in other limbs at different angles would then be tricky as you'd need separate frames ? Sorry if I've got completely the wrong end of the stick.

 

[sunsetpearl]

If you do wish to do the trig, I think you have the correct concept. (Parallel axis theorem, with thin rods) but you need to check extremes of the equation you develop to see over what angles it holds and what angles it may break down.

Write the equation for $m_2$ using the diagram. The red arrow is the distance $x$ of the parallel axis from the axis of rotation of rod $m_2$.

View attachment 339449

Thanks! Before I was attempting to use the parallel axis theorem with the parallel axis at the elbow joint at the other end of rod $m_2$. Why do we choose this way? My gut instinct without calculating is that it should yield the same result either way.

When I work this out I get $I$ = $1/3$$m_2$$l_2^2$$sin^2(π/2-β+α)$ + $m_2$($R$ + $l_1$$cos(α)$ - $l^2$$cos(β-α)$)²

Is it possible to do the same thing if the arm segments were represented as cylinders instead of rods? If not is there a better way you would suggest calculating (I suppose without/with less trig?).

 

[erobz]

Thanks! Before I was attempting to use the parallel axis theorem with the parallel axis at the elbow joint at the other end of rod $m_2$. Why do we choose this way? My gut instinct without calculating is that it should yield the same result either way.

When I work this out I get $I$ = $m_2$$/3$$l_2^2$$sin^2(π/2-β+α)$ + $m_2$($R$ + $l_1$$cos(α)$ - $l^2$$cos(β-α)$)²

Using my diagram variables, I get:

$$ I_{m_2} = m_2 \left(R + \ell_1 \cos \alpha - \ell_2 \cos ( \beta - \alpha ) \right)^2 + \frac{1}{3} m_2 {\ell_2}^2 \cos^2 ( \beta - \alpha) $$

But I'd say its only valid for $\beta - \alpha < \frac{\pi}{2}$

Is it possible to do the same thing if the arm segments were represented as cylinders instead of rods?

I believe what is being done is only valid for $\ell \gg r$. We are using the thin rod MOI approximation $\frac{1}{3}m \ell^2$

If not is there a better way you would suggest calculating (I suppose without/with less trig?).

I think not an easy way...

 

[jbriggs444]

View attachment 339434
I didn't include both arms in the diagram, but I assume the other arm would also be at some angle to the torso, not necessarily the same angle as the arm pictured.

The axis of rotation will not be through the center of the torso. It will be about a vertical line about which the arm(s) and the torso are in balance.

 

[erobz]

The axis of rotation will not be through the center of the torso. It will be about a vertical line about which the arm(s) and the torso are in balance.

Unless the arm configurations are symmetric across the torso, correct?

 

[jbriggs444]

Unless the arm configurations are symmetric across the torso, correct?

Yes, in that case, the "in balance" axis would correspond with the "center of torso" axis.